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Thank you, guys!  However, I tried $lastid = mysql_insertid() but it only inserts in the second table.  When I try to use that in a third table the value ends up being a zero.  Is this variable valid for inserting into one table only? Thanks!

Thank you, guys!  However, I tried $lastid = mysql_insertid() but it only inserts in the second table.  When I try to use that in a third table the value ends up being a zero.  Is this variable valid for inserting into one table only? Thanks!

Can you post the relevant code?

Actually problem got worse.  The code below gives me the second last id, not the last one.

if (!$connProj)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("databasename", $connProj);
$sql="INSERT INTO Submitter (Name, SNL_ID,Org, Email) VALUES
('$name', '$snlid', '$org', '$email')";
$lastid = mysql_insert_id();
echo "last id: " . $lastid;
if (!mysql_query($sql,$connProj))
  {
  die('Error: ' . mysql_error());
  }

 

So after doing the insert the browser might show me 80 as the last id but in the database the last id is 81. I have checked it many times because I couldn't believe this could happen.  Thank you so much for trying to help.

Isn't that what I am doing i.e. executing the insert query before getting the last insert id?  I am at my wits end because the same code on another page gives a last id of zero the first time and the second last one after I click the enter button in the address bar to reload the page.  I am thoroughly confused.  I acknowledge I must be a bit dense but I thought mysql_insert_id() was supposed to give the last id.

I have wasted so much time of so many of you good folks here just because I didn't understand something so basic.  Thanks to all who tried to help and most especially to Pikachu who finally showed me the light.  Thank you, guys and my apologies for testing your patience.

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